Coordinate Geometry Class 9 Extra Questions Advanced Maths Chapter 4

Students can refer to the NCERT Class 9 Advanced Maths Solutions and Chapter 4 Coordinate Geometry Extra Questions and Answers whenever they need help with difficult questions.

Class 9 Coordinate Geometry Extra Questions

Coordinate Geometry Class 9 Short Question Answer

Question 1.
Determine whether the lines are parallel, perpendicular, or neither.
y = 3x + 2 and y = 3x – 5
Concept Applied: Slopes of parallel lines are equal.
Solution:
Slope of first line: m₁ = 3
Slope of second line: m₂ = 3
Since, m₁ = m₂
The lines are parallel.

Question 2.
Points A( 1, 2), B(5, 4) and C(7, k) are the vertices of a right-angled triangle, right-angled at B. Find the value of k.
Solution:
Since the triangle is right-angled at B, then AB ⊥ BC Therefore, the product of their slopes is -1.
Slope of AB = \(\frac{4 – 2}{5 – 1}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
Slope of BC = \(\frac{k – 4}{7 – 5}\) = \(\frac{k – 4}{2}\)
Use perpendicular condition
Slope of AB × Slope of BC = -1
\(\frac{1}{2}\) × (\(\frac{k – 4}{2}\)) = -1
\(\frac{k – 4}{2}\) = -1
k – 4 = -4
k = 0

Question 3.
Point P(a, -2) is reflected in the x-axis to form point Q. Point Q is then translated 3 units to the left and 5 units upwards to reach the point R(-4, b). Find the value of a + b.
Concept Applied: Change in coordinates during translation:
• Left/Right movement changes the x-coordinate
• Upward/Downward movement changes the y-coordinate
Solution:
Given point P(a, -2).
After reflection in the x-axis, the y-coordinate changes sign.
So, Q(a, 2)
Now translate Q:
3 units left ⇒ subtract 3 from x-coordinate
5 units up ⇒ add 5 to y-coordinate
Therefore, R(a – 3, 2 + 5) or R(a -3, 7)
But R = (-4, B)
So, a – 3 = -4
a = -1
and b = 7
Now, a + b = -1 + 7 = 6
Hence, the required value is 6.

Question 4.
A point P(x, y) is reflected across the y-axis to form point Q. Point Q is exactly 12 units to the left of point P. Find the x-coordinate of point P.
Solution:
When a point is reflected across the y-axis:
(x, y) → (-x, y)
So, the horizontal distance between P and Q is x – (-x) = 2x
Given that Q is 12 units to the left of P,
2x = 12
x = 6
Hence, the x-coordinate of point P is 6.

Question 5.
(i) A point P(a, b) is reflected on the Y-axis to P'(-3, 1). Write down the values of a and b.
(ii) P” is the image of P when reflected in the X-axis. Write down the coordinates of P”.
(iii) P”’ is the image of P when reflected in the x = 5.
Write down the coordinates of P”.
Concept Applied: Reflection in the y-axis changes the sign of the x-coordinate while the y-coordinate remains unchanged.
Reflection in the x-axis changes the sign of the y-coordinate while the x-coordinate remains unchanged.
Solution:
(i) P(a, b) = P(3, 1)
a = 3, b = 1

(ii) P”(3, – 1)

(iii) F” ‘(7, 1)

Question 6.
Use a graph paper for this question. Take 2 cm = 1 unit along both the axes.
(i) Plot the points A(0, 4), B(2, 2), C(5, 2) and D(4, 0), E(0, 0) is the origin.
(ii) Reflect B, C, D on the Y-axis and name them as B’, C and D’ respectively.
(iii) Join the points ABCDD’C’B’ and A in order and give a geometrical name to the closed figure.
Solution:
(i) See graph,

(ii) See graph

(iii) On joining the points ABCDD’C’B’A, the shape obtained is not proper geometrical shape, but combination of a quadrilateral and triangle.

Question 7.
A straight line l has a slope -3. Another straight line m passes through the points P(2b, 1) and Q(b, 4). The lines l and m intersect at a right angle. Calculate the value of b and find the coordinates of points P and Q.
Concept Applied: Finding the slope of one line using the negative reciprocal of the other line’s slope.
Solution:
Given slope of line l,
m₁ = -3
Since, lines l and m are perpendicular,
m₁ × m₂ = -1
-3 × m₂ = -1
m₂ = \(\frac{1}{3}\)
Now find the slope of line m using points P(2b, 1) and Q(b, 4):
m₂ = \(\frac{4 – 1}{b – 2b}\) = \(\frac{3}{-b}\) = –\(\frac{3}{b}\)
Since slope is \(\frac{1}{3}\).
–\(\frac{3}{b}\) = \(\frac{1}{3}\)
b = -9
Coordinates of the points. Point P: (2b, 1) = (2 × -9, 1) = (-18, 1)
Point Q: (b, 4) = (-9, 4)
Hence, the coordinates are P(-18, 1) and Q(-9, 4).

Question 8.
Line p passes through the points P(a, 2) and Q(6, 8). Line q passes through the points R( 1, -3) and S(5, 5). If line p is parallel to line q, find the value of a. Hence, evaluate a² – 4.
Solution:
Since the lines are parallel, their slopes will be equal. Find slope of line q,
m_q = \(\frac{[5-(-3)]}{5-1}\) = \(\frac{8}{4}\) = 2
Find slope of line p,
m_p = \(\frac{8 – 2}{6 – a}\) = \(\frac{6}{6 – a}\)
Since, m_p = m_q
Therefore, \(\frac{6}{6 – a}\) = 2
6 = 12 – 2a
-6 = -2a
a = 3
Now, calculate, a² – 4 = (3)² – 4 = 9 – 4 = 5

Question 9.
Point P(2m – 9, m – 7) lies strictly in the quadrant III. The perpendicular distance of point P from the x-axis is twice its perpendicular distance from the y-axis. Find the value of m and hence evaluate m² + 6.
Concept Applied: Sign of coordinates in the III quadrant (-, -).
Solution:
Coordinates of point P are (2m – 9, m – 7)
Since the point lies in the III quadrant, so both x-coordinate and y – coordinate are negative.
So, 2m – 9 < 0 and m – 7 < 0
Distance from the x-axis: |m – 7|
Distance from the y-axis: |2m – 9|
Given: |m – 7| = 2|2m – 9|
Since both coordinates are negative, |m – 7| = 7 – m and |2m – 9| = 9 – 2m
Substituting, 7 – m = 2(9 – 2m)
7 – m = 18 – 4m
3m = 11
m = \(\frac{11}{3}\)
Verify quadrant condition
2m – 9 = 2(\(\frac{11}{3}\)) – 9
= \(\frac{22}{3}\) – \(\frac{27}{3}\) = – \(\frac{5}{3}\) < 0
m – 7 = \(\frac{11}{3}\) – 7
= \(\frac{(11 – 21)}{3}\)
= –\(\frac{10}{3}\) < 0
Hence, the point lies in the III quadrant.
Now, evaluate m² + 6 = (\(\frac{11}{3}\))² + 6
= \(\frac{121}{9}\) + 6
= \(\frac{175}{9}\)

Question 10.
The coefficients of the line Ax + By + C = 0 satisfy the relation A + 3B – 2C = 0. Show that the line always passes through a fixed point and find that point.
Solution:
Given line: Ax + By + C = 0
Condition: A + 3B – 2C = 0
A + 3B – 2C = 0
⇒ 2C = A + 3B
⇒ C = \(\frac{A+3 B}{2}\)
Substitute in the given line:
Ax + By + \(\frac{A+3 B}{2}\) = 0
Multiply by 2:
2Ax + 2By + A + 3B = 0
A(2x + 1) + B(2y + 3) = 0
Since A and B are arbitrary, therefore 2x + 1 = 0, 2y + 3 = 0
x = –\(\frac{1}{2}\), y = –\(\frac{3}{2}\)
Hence, the line always passes through the fixed point (-\(\frac{1}{2}\), –\(\frac{3}{2}\)).

Question 11.
Convert the linear equation 5x – 3y + 15 = 0 into slope-intercept form.
Concept Applied: Algebraic rearrangement of terms to isolate y.
Solution:
The given equation 5x – 3y + 15 = 0 can be written as 3y = 5x + 15
Thus, y = \(\frac{5 x}{3}\) + \(\frac{15}{3}\)
y = \(\frac{5 x}{3}\) + 5, is the required slope intercept form.

Question 12.
A straight line passes through the point (2, 3). The x-intercept a and y-intercept b of the line satisfy a + b = 0. Find all possible equations of the line. E
Solution:
Case I: The intercepts are non-zero (a, b ≠ 0)
Let the x-intercept be a and y-intercept be b. Since, a + b = 0, b = -a.
Use intercept form of line: \(\frac{x}{3}\) + \(\frac{15}{3}\) = 1
Put b = -a
Then, \(\frac{x}{a}\) + \(\frac{y}{-a}\) = 1
x – y = a
Use point (2, 3) in the equation, we get
2 – 3 = a
a = -1
Required equation is:
x – y = -1 or x – y + 1 = 0
Case II: Line passes through the origin
Then a = 0, b = 0, and a + b = 0.
The line through (0, 0) and (2, 3) is = \(\frac{y}{x}\) = \(\frac{3}{2}\)
3x – 2y = 0

Question 13.
A straight line passes through the points (2, 7) and (6, -1). Determine the equation of the line in slope- intercept form.
Concept Applied: The slope of the line passing through points is given by: \(\frac{y_2-y_1}{x_2-x_1}\)
Solution:
Let the equation of the required line be: y = mx + c.
slope, m = \(\frac{-1-7}{6-2}\) = –\(\frac{8}{4}\) = -2
So, the equation becomes: y = -2x + c
Since the line passes through (2, 7):
7 = – 2(2) + c
c = 11
Then, required equation of line is: y = – 2x + 11

Question 14.
Find the equation of the straight line perpendicular to the line x + 2y = 4, which cuts an intercept of 2 units from the positive y-axis. Hence, find the intersection point of the two lines.
Solution:
Given: x + 2y = 4 …….. (i)
On solving equation in terms of ‘y’ we get,
2y = – x + 4
⇒ y = –\(\frac{1}{2}\)x + 4
On comparing above equation with y = mx + c we get,
m = –\(\frac{1}{2}\)
Let, slope of the straight line perpendicular to the line x + 2y = 4 be m₁.
⇒ m₁ × –\(\frac{1}{2}\) = -1
⇒ m₁ = 2
Now, on substituting value of m₁ in y = m₁x + 2(as perpendicular lines cuts the intercept 2 units from the positive y-axis)
we get,
y = 2x + 2 ……….. (ii)
On substituting value of ‘y’ in equation (i) we get,
x + 2(2x + 2) = 4
⇒ x + 4x + 4 = 4
⇒ 5x = 0
⇒ x = 0
Now substituting value of ‘x’ in equation (ii) we get,
y = 2 × 0 + 2
= 0 + 2 = 2.
Hence, the equation of the straight line is y = 2x + 2 and the intersection point of the two lines is (0, 2).

Question 15.
Find the equation of a line that is parallel to the line y = 3x – 5 and whose y-intercept is –\(\frac{1}{5}\).
Concept Applied: Parallel lines have equal slopes.
Solution:
It is given that the y-intercept (c) of the required line is = –\(\frac{1}{5}\).
The equation of the given line is, y = 3x – 5
Comparing this with y = mx + b, we get its slope to be m = 3.
Since the given line is parallel to the required line, their slopes are equal.
So the slope of the required line is, m’ = 3 as well.
Thus the equation of the required line using the slope- intercept formula is,
y = m’x + C
y = 3x – \(\frac{1}{5}\)
The equation of the required line is, y = 3x – \(\frac{1}{5}\).

Question 16.
A straight line passes through the point (4, 3) and forms a right-angled triangle with the positive x-axis and y-axis. If the area of the triangle is 24 square units, find the values of its x-intercept (a) and y-intercept (b).
Solution:
Let the line cut the axes at:
x-intercept = a
y-intercept = b
So the equation of the line is:
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Since the line passes through (4,3): \(\frac{4}{a}\) + \(\frac{3}{b}\) = 1 ………. (i)
Use area of triangle
Area = \(\frac{1}{2}\) ab = 24
ab = 48
So, b = \(\frac{48}{a}\)
Substituting the value in equation (i), we get
\(\frac{4}{a}\) + \(\frac{\frac{3}{48}}{a}\) = 1
\(\frac{4}{a}\) + \(\frac{a}{16}\) = 1
64 + a² = 16a
a² – 16a + 64 = 0
(a – 8)² = 0
a = 8
Then, b = \(\frac{48}{8}\) = 6
Hence, a = 8 and b = 6.

Question 17.
A straight line passes through the point (2, -1) and has slope \(\frac{3}{2}\). The line forms an x-intercept at point A and a y-intercept at point B.
(i) Find the equation of the line in slope-intercept form.
(ii) Find the y-intercept.
(iii) Find the x-intercept.
Solution:
(i) Equation of line
Using slope-intercept form:
y = mx + c
Given m = –\(\frac{3}{2}\), so
y = –\(\frac{3}{2}\)x + c
Substitute point (2, -1), we get
-1 = –\(\frac{3}{2}\) × 2 + c
c = 2
So, the required equation is:
y = –\(\frac{3}{2}\)x + 2

(ii) y-intercept
At x = 0, y = 2
So y-intercept is: (0, 2)

(iii) x-intercept
Put y = 0:
0 = –\(\frac{3}{2}\)x + 2
x = \(\frac{4}{3}\)
So x-intercept is: (\(\frac{4}{3}\), 0)

Question 18.
In the given figure, line l₁, has a slope defined by its vertical rise and horizontal run, where the rise and run are equal. Line l₁ is parallel to line l₂. If line l₃ is perpendicular to line l₁ then find the slopes of lines l₂ and l₃.

Solution:
Given that:
For line l₁, the rise and run are equal.
So, Slope of line l₁ = 1
We know that, if two lines are parallel, their slopes are equal.
Thus, slope of line l₂ = slope of line l₁ = 1
Also, if two lines are perpendicular, product of their slopes = -1
⇒ Slope of line l₂ × Slope of line l₃ = -1
⇒ 1 × Slope of line l₃ = -1
Thus, slope of line l₃ = -1

Question 19.
Find the equation of the straight line perpendicular to the line 3x + 4y = 12, which cuts an intercept of 3 units on the positive x-axis.
Hence, find:
(i) The equation of the required line
(ii) The point of intersection of the two lines
(iii) The intercept it makes on the y-axis
Solution:
(i) Given line: 3x + 4y = 12
Rewrite in slope form:
4y = – 3x + 12
⇒ y = –\(\frac{3}{4}\)x + 3
So, slope m₁ = –\(\frac{3}{4}\)
Slope of perpendicular line: m₂ = \(\frac{3}{4}\) [For perpendicular lines: m₁ × m₂ = -1]
The line cuts an intercept of 3 units on positive x-axis, so it passes through (3, 0)
Using point-slope form:
y – 0 = \(\frac{4}{3}\)(x – 3)
y = \(\frac{4}{3}\)x – 4
Required equation of line is y = \(\frac{4}{3}\)x – 4

(ii) Now, substitute y = \(\frac{4}{3}\)x – 4 in 3x + 4y = 12
3x + 4(\(\frac{4}{3}\)x – 4) = 12
3x + \(\frac{16}{3}\)x – 16 = 12
\(\frac{25}{3}\)x = 28
x = \(\frac{84}{25}\)
Now find y:
y = \(\frac{4}{3}\) . \(\frac{84}{25}\)
= \(\frac{112}{25}\) – 4
= \(\frac{112 – 100}{25}\) = \(\frac{12}{25}\)
Point of intersection is (\(\frac{84}{25}\), \(\frac{12}{25}\))

(iii) For y-iritercept of required line
Put x = 0:
y = -4
y-intercept is (0, -4).

Coordinate Geometry Class 9 Long Question Answer

Question 1.
Use a graph paper for this question (Take 2 cm = 1 unit on both X and Y-axis)
(i) Plot following points: A(0, 4), B(2, 3), C(1, 1) and D(2, 0).
(ii) Reflect points B, C, D on the Y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
(iii) Join the points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation of the line of symmetry of the figure formed.
Solution:
(i)

(ii) Coordinates of images of B, C and D are B'(-2, 3), C'(-1, 1) and D'(-2, 0).

(iii) Equation of line of symmetry is x = 0, i.e., the y-axis

Question 2.
Point P(3m + 1, m – 5) lies strictly in the quadrant IV. The perpendicular distance of point P from the x-axis is three times its perpendicular distance from the y-axis. Find the value of m and hence evaluate m² + 8.
Concept Applied: The perpendicular distance of a point from the x-axis is the absolute value of its y-coordinate, while the perpendicular distance from the y-axis is the absolute value of its x-coordinate.
Solution:
Coordinates of point P are: (3m + 1, m – 5)
Since the point lies in the IV quadrant:
• x-coordinate is positive
• y-coordinate is negative
So, 3m + 1 > 0 and m – 5 < 0
Distance from the x-axis = |m – 5|
Distance from the y-axis = |3m + 1|
Given: |m – 5| = 3|3m + 1|
Since the point is in IV quadrant: m – 5 < 0 and 3m + 1 > 0
Therefore, |m – 5| = 5 – m
And |3m + 11 = 3m + 1
Substituting:
5 – m = 3(3m + 1)
5 – m = 9m + 3
5 – 3 = 10m
2 = 10m
m = \(\frac{1}{5}\)
3m + 1 = \(\frac{3}{5}\) + 1 = \(\frac{8}{5}\) > 0
m – 5 = \(\frac{1}{5}\) – 5 = –\(\frac{24}{5}\) < 0
So the point lies strictly in Quadrant IV
Now, evaluate:
m² + 8 = (\(\frac{1}{5}\))² + 8 = \(\frac{1}{25}\) + 8
= \(\frac{1 + 200}{25}\) = \(\frac{201}{25}\)

Question 3.
A straight line l passes through the point (0, 1) and intersects the line segment joining the points A(2, 7) and B(6, 3) at a point P.
(i) Find the slope of the line joining (0, 1) and A.
(ii) Find the slope of the line joining (0, 1) and B.
(iii) Hence, determine the possible range of values of the slope m of line l.
(iv) State whether the line with slope m = 4 can intersect the segment AB. Give reason.
Solution:
(i) Slope of line joining (0, 1) and A(2, 7) is:
m_A = \(\frac{7 – 1}{2 – 0}\) = \(\frac{6}{2}\) = 3
So, the slope is 3.

(ii) the slope of the line joining (0, 1) and B(6, 3) is:
m_B = \(\frac{3 – 1}{6 – 0}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Hence, the slope is \(\frac{1}{3}\)

(iii) Since line l intersects the segment joining A and B, its slope must lie between the slopes formed with points A and B.
Therefore, \(\frac{1}{3}\) ≤ m ≤ 3.
Hence, the required range is \(\frac{1}{3}\) ≤ m ≤ 3.

(iv) Given, m = 4
But, 4 > 3
So, 4 does not lie in the required range.
Therefore, a line with slope 4 cannot intersect the segment AB.
Hence, the answer is No.

Question 4.
Line p₁ has the equation 2x + 7y – 14 = 0. Line p₂ has the equation 7x – 2y + C = 0.
(i) Prove algebraically that the lines p₁ and p₂ are perpendicular.
(ii) If the y-intercept of p₁ is equal to the x-intercept of p₂, find the value of C.
Solution:
(i) Slope of p₁
Given: 2x + 7y – 14 =0
Rewrite in slope-intercept form: 7y = – 2x + 14
y = \(\frac{2 x}{7}\) + 2
So, slope of p₁ is m₁ = –\(\frac{2}{7}\).
Slope of p₂
Given: 7x – 2y + C = 0
– 2y = – 7x – C
y = \(\frac{7 x}{2}\) + \(\frac{C}{2}\)
So, slope of p₂ is m₂ = \(\frac{7}{2}\)
Then, m₁ × m₂ = –\(\frac{2}{7}\) × \(\frac{7}{2}\) = -1
Hence, the lines are perpendicular.

(ii) y-intercept of p₁
Put x = 0:
7y – 14 = 0
y = 2
So, y-intercept of p₁ is 2.
x-intercept of p₂
Put y = 0:
7x + C = 0
x = –\(\frac{C}{2}\)
Given that both intercepts are equal:
–\(\frac{C}{7}\) = 2
C = -14
Hence, the required value is -14.

Question 5.
Find the equation of the line passing through the point of intersection of the line 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Solution:
Let the equation of the line having equal intercepts on the axes be:
\(\frac{x}{a}\) + \(\frac{y}{a}\) = 1
x + y = a …….. (i)
On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0
we obtain x = \(\frac{1}{13}\) and y = \(\frac{5}{13}\).
Therefore, (\(\frac{1}{13}\), \(\frac{5}{13}\)) is the point of the intersection of the two given lines.
Since equation (i) passes through the point (\(\frac{1}{13}\), \(\frac{5}{13}\))
\(\frac{1}{13}\) + \(\frac{5}{13}\) = a
\(\frac{6}{13}\) = a
Equation (i) becomes x + y = \(\frac{6}{13}\), i.e., 13x + 13y = 6
Thus, the required equation of the line is 13x + 13y = 6.

Question 6.
Given the equations of two straight lines, L₁ and L₂ are x – y = 1 and x + y = 5 respectively. If L₁ and L₂ intersects at point Q(3, 2). Find:
(a) the equation of line L₃ which is parallel to L₁ and has y-intercept 3.
(b) the value of k, if the line L₃ meets the line L₂ at a point
P(k, 4).
Solution:
(a) Given: L₁ as x – y = 1 ⇒ y = x – 1
On comparing it with y = mx + c we get, m = 1
As, slope of parallel lines are equal
Slope of L₃ = 1
As L₃ has y-intercept = 3
y = mx + c
⇒ y = 1.x + 3
⇒ y = x + 3
Hence, equation of line L₃ is y = x + 3

(b) As, L₂ is x + y = 5 and L₃ is y = x + 3
On substituting the value of y in x + y = 5, we get,
x + (x + 3) = 5
⇒ 2x + 3 = 5
x = \(\frac{1}{2}\)
⇒ x = 1
Now, substitute value of x in y = x + 3 we get,
y = 1 + 3 = 4
⇒ (x, y) = (1, 4)
Hence, value of k = 1

Coordinate Geometry Class 9 Case Based Questions

1. A city is designed using a Cartesian coordinate system to plan roads efficiently. The main highway is represented by the line:
y = \(\frac{2}{3}\)x – 4
Engineers are designing new roads that must follow rules of slope, intercepts and geometry of lines.
• Road A is perpendicular to the main highway.
• Road B is parallel to the main highway and passes through the point (6, 2).
• Road C passes through (0, 5) and (3, 0).

Question 1.
What is the slope of Road A (perpendicular to the main highway)?
(a) \(\frac{2}{3}\)
(b) –\(\frac{3}{2}\)
(c) \(\frac{3}{2}\)
(d) –\(\frac{2}{3}\)
Answer:
Option (b) is correct
Explanation: Slope of main road = \(\frac{2}{3}\)
Perpendicular slope = negative reciprocal:
m = –\(\frac{3}{2}\)

Question 2.
What is the equation of Road B?
(a) y = \(\frac{2 x}{3}\) – 2
(b) y = \(\frac{2 x}{3}\) + 2
(c) y = \(\frac{-2 x}{3}\) + 2
(d) y = \(\frac{3 x}{3}\) – 2
Answer:
Option (a) is correct
Explanation: Since, Road B is parallel to the main highway, so it has same slope, i.e. m = \(\frac{2}{3}\)
So, equation of line is: y = mx + c Or y = \(\frac{2 x}{3}\) + c
Also, it passes through point (6, 2):
2 = \(\frac{2}{3}\) × 6 + c
2 = 4 + c
c = -2
So, required equation is: y = \(\frac{2 x}{3}\) – 2

Question 3.
What is the x-intercept of Road C?
(a) 3
(b) 5
(c) 0
(d) -3
Answer:
Option (a) is correct
Explanation: x-intercept is where y = 0.
Road C passes through (0, 5) and (3, 0) already meets both axes at those points.
At x = 3, y = 0 → this is the point where it cuts the x-axis.
So x-intercept = 3 units

Question 4.
What is the slope of Road C?
(a) –\(\frac{5}{3}\)
(b) \(\frac{5}{3}\)
(c) –\(\frac{3}{5}\)
(d) \(\frac{3}{5}\)
Answer:
Option (a) is correct
Explanation: Road C passes through (0, 5) and (3, 0), therefore, slope, m = \(\frac{0 – 5}{3 – 0}\) = –\(\frac{5}{3}\)

Question 5.
Which of the following statements is correct about the main highway?
(a) It has negative slope
(b) It is horizontal
(c) It has positive slope
(d) It is vertical
Answer:
Option (c) is correct
Explanation: y = \(\frac{2 x}{3}\) – 4
Slope = \(\frac{2}{3}\) = positive ⇒ line rises left to right

2. A city metro system is planned using a Cartesian coordinate grid. The main metro line is represented by the equation:
y = –\(\frac{1}{2}\) x + 6
Engineers design connecting routes based on slope and intercept properties.
Line A is perpendicular to the main metro line.
Line B is parallel to the main metro line and passes through the point (4, 3).
Line C passes through the points (2, 4) and (6, 0).

Question 1.
Find the slope of Line A.
Solution:
Slope of main metro line = –\(\frac{1}{2}\)
Perpendicular slope = negative reciprocal:
m = 2
Thus, slope of line A is 2.

Question 2.
Find the slope of line B.
Solution:
Parallel lines have equal slope.
So, slope of Line B = slope of main metro line,
i.e., m = \(\frac{1}{2}\)

Question 3.
Find the equation of Line B.
Solution:
Slope of line B is –\(\frac{1}{2}\)
So, equation of line B in point-slope form can be:
y = –\(\frac{x}{2}\) + c
As this line passes through the point (4, 3), it satisfies the given equation.
3 = –\(\frac{4}{2}\)
3 = – 2 + c
c = 5
Then, equation of line is : y = –\(\frac{x}{2}\) + 5
Or
2y + x – 10 = 0
OR
Question 3.
Find the equation of Line C.
Solution:
Slope of line C = \(\frac{0 – 4}{6 – 2}\) = –\(\frac{4}{4}\) = -1
So, the equation of line C in point-slope form will be: y = -x + c
The line passes through point (2, 4), then:
4 = – 2 + c
c = 6
So, the equation of required line is: y + x – 6 = 0

Coordinate Geometry Class 9 Extra Questions Very Short Answer Type

Question 1.
Write the signs convention of the coordinates of a point in the second quadrant.
Solution:
(-ve, +ve)

Question 2.
Write the value of ordinate of all the points lie on x-axis.
Solution:
0

Question 3.
Write the value of abscissa of all the points lie on y-axis.
Solution:
0

Question 4.
If in coordinates of a point B(3, -2), signs of both coordinates are interchanged, then it will lie in which quadrant ?
Solution:
When signs of both coordinates of B(3, -2) are interchanged, then coordinates of new point are B'(-3, 2) and it will lie in second quadrant.

Question 5.
Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis.
Solution:
Distances of point C(-3, -2) from x-axis is 2 units in the negative direction and from y-axis is 3 units in the negative direction. Distances of point D(5, 2) from x-axis is 2 units and from y-axis is 5 units.

Question 6.
Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal.
Solution:
Here, two ordered pairs are equal.
⇒ Their first components are equal and their second components are separately equal.
⇒ x – 3 = 4 and x + y = -6
⇒ x = 7 and 7 + y = -6 ⇒ y = – 13
Hence, x = 7 and y = – 13.

Question 7.
In which quadrant does the point (-1, 2) lie ?
Solution:
(-1, 2) lie in second quadrant.

Question 8.
Find the distance of the point (0, -5) from the origin.
Solution:
5 units.

Question 9.
Write the shape of the quadrilateral formed by joining (1, 1), (6, 1), (4, 5) and (3, 5) on graph paper.
Solution:
Trapezium.

Coordinate Geometry Class 9 Extra Questions Short Answer Type 1

Question 1.
In the given figure, ABCD is a rectangle with length 6 cm and breadth 3 cm. O is the mid-point of AB. Find the coordinates of A, B, C and D.

Solution:

We have taken 1 cm = 1 unit and origin O is the mid-point of AB
∴ OA = OB = 3 cm
and BC = AD = 3 cm
Thus, the coordinates of A are (-3, 0)
the coordinates of B are (3, 0)
the coordinates of C are (3, 3)
the coordinates of D are (-3, 3)

Question 2.
Write the coordinates of A, B, C and D from the figure given alongside.

Solution:
Coordinates of the point A are (5, 0)
Coordinates of the point B are (5, 3)
Coordinates of the point C are (-2, 4)
Coordinates of the point D are (0, -2)

Question 3.
A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates? What will be the coordinates of a point, if it lies on y-axis at a distance of -9 units from x-axis ?
Solution:
As shown in graph, the coordinates of a point which lies on x-axis at a distance of 9 units from y-axis are (9, 0) and the coordinates of a point which lies at a distance of -9 units from x-axis are (0, -9).

Question 4.
Plot the point P(2, -6) on a graph paper and from it draw PM and PN perpendiculars to x-axis and y-axis respectively. Write the coordinates of the points M and N.
Solution:

As shown in graph, coordinates of M are (2, 0) and coordinates of N are (0, -6).

Coordinate Geometry Class 9 Extra Questions Short Answer Type 2

Question 1.
Without plotting the points indicate the quadrant in which they lie, if :
(i) ordinate is 5 and abscissa is – 3
(ii) abscissa is -5 and ordinate is – 3
(iii) abscissa is – 5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Solution:
(i) Clearly, point (-3, 5) lies in 2nd quadrant.
(ii) Clearly, point (-5, – 3) lies in 3rd quadrant.
(ii) Clearly, point (-5, 3) lies in 2nd quadrant.
(iv) Clearly, point (3, 5) lies in 1st quadrant.

Question 2.
Plot the points A(1, 4), B(-2, 1) and C(4, 1). Name the figure so obtained on joining them in order and also, find its area.
Solution:

Triangle.
Area of ∆ABC = \(\frac{1}{2}\) × BC × Height
= \(\frac{1}{2}\) × 6 × 3
= 9 sq. units

Question 3.
Plot the following points, join them in order and identify the figure thus formed : A(1, 3), B(1, -1), C(7, -1) and D(7, 3)
Write the coordinates of the point of intersection of the diagonals.
Solution:

ABCD is a rectangle.
Point of intersection of the diagonals AC and BD is (4, 1).

Question 4.
Plot the points A(2, 5), B(8,5) and C(5, -3) and join AB, BC and CA. What figure do you obtain ?
Solution:

We obtain an isosceles triangle in which AC = BC.

Question 5.
(i) Plot the points M(4, 3), N(4, 0), 0(0, 0), P(0, 3).
(ii) Name the figure obtained by joining MNOP.
(iii) Find the perimeter of the figure.
Solution:
(i)

(ii) As shown in graph, the figure obtained by joining MNOP is rectangle.
(iii) Perimeter of rectangle MNOP = 2 (ON + OP) = 2 (4 + 3) = 2 × 7 = 14 units.

Question 6.
Plot D(-2, -3) on the graph paper. Also, plot reflections of D in x-axis and y-axis.
Solution:

Reflection of D(-2, -3) in x-axis is D'(-2, 3) and reflection of D(-2, -3) in y-axis is D'(2, -3).

Question 7.
If the coordinates of a point M are (-2, 9) which can also be expressed as (1 + x, y) and y > 0, then find in which quadrant do the following points lie : Ply, x), Q(2, x), R(x, y − 1), S(2x, -3y).
Solution:
Here, M(-2, 9) can also be expressed as (1 + x, y²)
∴ (1 + x, y²) ⇔ (-2, 9) and y > 0
⇒ 1 + x = -2 and y² = 9
⇒ x = -3 and y = √9 = 3 (∵ y > 0)
P (y, x) = P(3, -3), it lies in IV quadrant
Q(2, x) = Q(2, -3), also lies in IV quadrant
R(x², y – 1) = R((-3)², 3 – 1) = R(9, 2), it lies in I quadrant
S(2x, -3y) = S(2 × (-3), -3 × 3) = S(-6, -9), it lies in III quadrant

Question 8.
In the given figure, PQR is an equilateral triangle with coordinates of Q and R as (-2, 0) and (2, 0) respectively. Find the coordinates of the vertex P.
Solution:

Here,
QR = OQ + OR.
= 2 + 2 = 4 units
∴ ∆PQR is an equilateral triangle.
∴ PQ = PR = QR = 4 units
In right-angled ∆OPQ, ∠POQ = 90°
∴ We have PQ² = OP² + OQ²
⇒ OP² = PQ² – OQ²
= 4² – 2² = 16 – 4 = 12
⇒ OP = √12 = 2√3 units
∴ The coordinates of P are (0, 2√3).

Coordinate Geometry Class 9 Extra Questions HOTS

Question 1.
Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed.
Solution:

Figure so formed is ABCD a rectangle.

Question 2.
Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed.
Solution:
Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance ‘a’on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance ‘a’ on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph. Now, PQRS, thus formed is a square.

Coordinate Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
On environment day, class-9 students got five plants of mango, silver oak, orange, banyan and amla from soil department. Students planted the plants and noted their locations as (x, y).

Plot the points (x, y) in the graph and join them in the given order. Name the figure you get. Which social act is being done by students of class-9 ?
Solution:

The given trees (points) are Mango (2, 0), Silver Oak (3, 4), Orange (0,7), Banyan (-3, 4) and Amla (-2, 0). The location of these trees are Orange (0,7) shown in the graph.
On joining the points of mango, silver oak, orange, banyan and amla in order, the figure so formed is a regular pentagon.
Planting more trees helpful in reducing pollution and make the environment clean and green for the coming generations.